- WELCOME TO A LESSON ON THE COUNTING PRINCIPLE.
AND THE GOAL OF THIS VIDEO IS TO DETERMINE THE NUMBER OF WAYS
MULTIPLE AND INDEPENDENT EVENTS CAN OCCUR.
SO THIS IS WHAT THE FUNDAMENTAL COUNTING PRINCIPLE SAYS:
IF THERE IS A SEQUENCE OF INDEPENDENT EVENTS
THAT CAN OCCUR, "A" SUB 1, "A" SUB 2, ALL THE WAY OUT TO "A"
SUB N WAYS, THEN THE NUMBER OF WAYS ALL OF THE EVENTS
CAN OCCUR IS "A" SUB 1 x "A"
SUB 2 x ALL THE WAY OUT TO "A" SUB N.
SO WE CAN USE THE COUNTING PRINCIPLE TO ANSWER A QUESTION,
AS WE SEE HERE.
HOW MANY WAYS CAN STUDENTS ANSWER A THREE-QUESTION
TRUE OR FALSE QUIZ?
BEFORE WE TAKE A LOOK AT THE COUNTING PRINCIPLE THOUGH,
LET'S GO AHEAD AND ILLUSTRATE ALL OF THE POSSIBLE WAYS
TO ANSWER THIS TEST USING A TREE DIAGRAM.
SO FOR QUESTION ONE THERE ARE TWO CHOICES.
IT MUST BE ANSWERED TRUE OR FALSE FOR QUESTION ONE.
NOW, FOR QUESTION TWO THERE'S ALSO TWO OPTION, TRUE OR FALSE,
BUT SINCE THERE'S ALSO TWO WAYS TO ANSWER THE FIRST QUESTION,
IF THE FIRST QUESTION IS TRUE,
THE SECOND QUESTION COULD BE TRUE OR FALSE.
AND ALSO, IF THE FIRST QUESTION IS FALSE,
WE HAVE THE OPTION OF TRUE OR FALSE FOR THE SECOND QUESTION.
SO WE CAN SEE AFTER QUESTION TWO WE HAVE FOUR POSSIBILITIES,
TRUE/TRUE, TRUE/FALSE, FALSE/TRUE, OR FALSE/FALSE,
EACH OF THOSE BEING UNIQUE ANSWERS TO THIS QUIZ.
AND THEN LASTLY, FOR THE THIRD QUESTION
THERE'S ALSO TWO OPTIONS,
SO AT THE END OF EACH BRANCH HERE THERE ARE TWO OPTIONS.
SO WE HAVE TRUE/FALSE HERE, TRUE/FALSE HERE, HERE,
AND HERE AS WELL.
SO EACH OF THESE BRANCHES REPRESENTS A POSSIBLE OPTION
FOR ANSWERING THE QUESTIONS ONE, TWO, AND THREE FOR THIS QUIZ,
AND YOU CAN SEE THERE ARE EIGHT DIFFERENT WAYS:
TRUE/TRUE/TRUE, TRUE/TRUE/FALSE, TRUE/FALSE/TRUE,
TRUE/FALSE/FALSE, AND SO ON ALL THE WAY DOWN THIS TREE.
NOW, IT'S MUCH FASTER JUST TO USE THE COUNTING PRINCIPLE.
THERE ARE THREE EVENTS TAKING PLACE HERE,
ONE FOR EACH QUESTION.
AND EACH OF THESE QUESTIONS CAN BE ANSWERED TWO DIFFERENT WAYS.
SO USING THE COUNTING PRINCIPLE WE'D HAVE 2 x 2 x 2
FOR THE FIRST THREE QUESTIONS,
WHICH EQUALS 8 POSSIBLE WAYS THIS QUIZ CAN BE ANSWERED.
LET'S TAKE A LOOK AT ANOTHER QUESTION.
HOW MANY PASSWORDS USING 6 DIGITS,
WHERE THE FIRST 2 DIGITS MUST BE LETTERS,
AND THE LAST 4 DIGITS MUST BE NUMBERS?
SO THERE ARE 6 INDEPENDENT EVENTS HERE, 1, 2, 3, 4, 5, 6.
THE FIRST TWO MUST BE LETTERS,
AND THE LAST FOUR MUST BE NUMBERS.
WELL, THERE ARE 26 LETTERS IN THE ALPHABET,
SO THERE'S 26 WAYS THE FIRST LETTER CAN BE CHOSEN,
AND THE SAME FOR THE SECOND LETTER.
AND THEN THE LAST 4 MUST BE NUMBERS.
WELL, THE DIGITS FROM 0 TO 9 REPRESENT 10 CHOICES
FOR EACH NUMBER, SO THIS WOULD BE x 10 x 10 x 10 x 10.
SO THIS PRODUCT WILL REPRESENT THE TOTAL NUMBER OF PASSWORDS
THAT ARE POSSIBLE USING THESE CONDITIONS.
SO LET'S GO AHEAD AND FIGURE OUT WHAT THAT WOULD BE.
SO IF WE USE OUR CALCULATOR WE WOULD HAVE 26 SQUARED x 10
TO THE 4th = 6,760,000 POSSIBLE PASSWORDS.
LET'S GO AHEAD AND TAKE A LOOK AT A COUPLE MORE QUESTIONS.
A RESTAURANT OFFERS A DINNER SPECIAL IN WHICH YOU GET TO PICK
ONE ITEM FROM FOUR DIFFERENT CATEGORIES,
AS WE SEE LISTED HERE.
HOW MANY DIFFERENT MEALS ARE POSSIBLE?
FIRST THING WE SHOULD NOTICE IS WE HAVE FOUR DIFFERENT
CATEGORIES TO SELECT FROM,
SO WE HAVE 1, 2, 3, 4 INDEPENDENT EVENTS.
SO THE FIRST EVENT WOULD BE FOR THE BEVERAGE,
THE SECOND EVENT WOULD BE FOR THE APPETIZER,
THIRD FOR THE MEAL, AND FOURTH FOR THE DESSERT.
LOOKING AT THE BEVERAGE OPTION
YOU CAN SEE THERE ARE FOUR POSSIBILITIES,
SO THERE ARE FOUR WAYS THAT EVENT CAN OCCUR.
FOR APPETIZERS THERE ARE THREE,
FOR THE MEAL THERE ARE 1, 2, 3, 4 OPTIONS
OR FOUR WAYS TO CHOOSE THE MAIN MEAL,
AND FOR THE DESSERT WE HAVE THREE OPTIONS.
SO THE PRODUCT OF THESE WILL TELL US THE TOTAL NUMBER OF WAYS
WE CAN SELECT DIFFERENT DINNERS FROM THIS SPECIAL.
WELL, 4 x 3 = 12 x 4 = 48 x 3 = 144,
SO THERE ARE 144 WAYS TO SELECT A DINNER UNDER THESE CONDITIONS.
LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE.
A DOOR LOCK ON A CLASSROOM REQUIRES ENTRY OF FOUR DIGITS.
ALL DIGITS MUST BE NUMBERS, BUT THE DIGITS CANNOT BE REPEATED.
HOW MANY UNIQUE CODES ARE POSSIBLE?
SO AGAIN, THIS IS VERY SIMILAR,
EXCEPT WE DO HAVE A SPECIAL CONDITION HERE
THAT THE DIGITS CANNOT REPEAT.
SO IT'S A SLIGHT VARIATION OF THE COUNTING PRINCIPLE,
SINCE EACH OF THESE DIGITS ARE NOT INDEPENDENT,
BECAUSE THEY CANNOT BE REPEATED.
HOWEVER, FOR THE FIRST DIGIT THERE ARE NO RESTRICTIONS,
SO THERE ARE 10 POSSIBILITIES, 0 THROUGH 9.
BUT FOR THE SECOND DIGIT WE CANNOT REPEAT THE DIGIT
USED IN THE FIRST POSITION,
SO INSTEAD OF 10 POSSIBILITIES THERE WOULD ONLY BE 9.
AND THEN FOR THE THIRD WE CAN'T REPEAT
EITHER OF THE FIRST TWO DIGITS,
SO THAT WOULD LEAVE 8 POSSIBILITIES.
AND THEN FOR THE FOURTH WE CAN'T REPEAT ANY OF THE FIRST THREE,
SO THAT WOULD LEAVE 7 POSSIBLE DIGITS.
SO THE PRODUCT OF THESE WOULD BE THE TOTAL NUMBER OF UNIQUE
CODES THAT ARE POSSIBLE.
SO WE HAVE 10 x 9 x 8 x 7 = 5,040 UNIQUE CODES
FOR THE DOOR LOCK.
OKAY, THAT WILL DO IT FOR THIS VIDEO.
NEXT WE'LL TAKE A LOOK AT PERMUTATIONS AND COMBINATIONS.
THANK YOU FOR WATCHING.